Polyatomic molecules - Cobalt

CHEMISTRY 373 - TUTORIAL 12
PRACTICE FOR POLYATOMIC MOLECULES AND VARIATION
THEORY
APRIL 5
1.
Write down the secular determinants for (a) linear H 3. (b) cyclic H 3 , within the Hüchel
approximation. By calculating the total energy as
E T = n 1ε1+ n 2ε2 + n3ε3
Discuss the relative stability of the two conformations (a) and (b) for
H 3 + -
; H 3 and H3.
Answer :
For (a) we get the secular determinant
α − E
β α−
E β = 0
β α−
E
sin ce the two terminal hydrogens H 1 and H3
do not interact
H 2
H 1
H 3
H 2
α−
E β
β α−
E β
β α−
E
=
H 1
H 3
α−
E β
( α − E) ( ) − β β β
β ( α−
E) 0 ( α − E)
3 2 2
= ( α−E) −β ( α−E) −β ( α−E)
2 2
= ( α−E) α−E)
−2β
0
[ ] =
ε3 = α−
2β
ε2 = α
Thus
ε1 = α+ 2β;
ε2 = α;
ε3
= α−
2β
ε1 = α+
2β

We have for conformation (a) the energies
H 3 + : 2 α +2 2β
H
H
3
3 −
: 3 α +2 2β
: 4 α +2 2β
For the second conformation (b)
α-E
β β
β α-E
β
β β α-E
=
ε2 = ε3
= α−β
α-E
β
-E
( α -E)
− β β β + β β α
β α-E β α-E
β β
3 + 3
ε1 = α+
2β
3 − : 4 α +2β
3 2 2 3
3 2 3
1 = + 2 ; 2 = 3 = −
3 + α β
3 : 3 α +3β
3 -
( α-E) −( α-E) β − 2β ( α-E)-2β
=
( α-E)
− 3(
α-E)
β + 2β
[( α-E)-β][ ( α-E)-β][ ( α-E)+2β]=
0
Thus
ε α β ε ε α β
We have for conformation (b) the energies
H : 2 +4
H
H
Thus H and H will prefer (b) but H
will prefer (a)

2. (a) Consider the allyl molecule CH 2CHCH2.
Construct
the secular determinant and find the orbital energies for the
π - electron system. Find the molecular orbitals and scetch them.
You can make use of the results from the previous question on H3
(b) Do the same for cyclopropene
H
C
H 2 C
CH 2 HC
H
C
CH
Answer :
For (a) we get the secular determinant
α − E
β α−
E β
β α−
E
= 0
sin ce the two terminal hydrogens C 1 and C3
do not interact
α−
E β
β α−
E β
β α−
E
=
α−
E β
( α − E) ( ) − β β β
β ( α−
E) 0 ( α − E)
3 2 2
= ( α−E) −β ( α−E) −β ( α−E)
2 2
= ( α−E) α−E)
−2β
0
[ ] =
Thus
ε1 = α+ 2β;
ε2 = α;
ε3
= α−
2β

The set of linear equations reads
C ( α− E)
+ C β = 0
11
12
C β + C ( α− E)
+ C β =0
11 12
13
C
12
β
+C
12
( α− E)
= 0
We get
1 1 1
1 1
1 1 1
ψ = ϕ + ϕ + ϕ ; ψ = ϕ − ϕ ; ψ = ϕ − ϕ + ϕ
2 2 2
2 2 2 2 2
1 1 2 3 2 1 3 3 1 2 3
3. (a) In the ' free electron molecular orbital' (FEMO) theory, the electrons
in a conjugated molecule are treated as independent particles in a box
of length L . Sketch the form of the two occupied orbitals in butadiene, predicted
by this model and predict the minimum excitation energy of the molecule.
Comments on the relation to the π - orbitals obtained in Atkins. What are the
similarities and the differences. What are the differences between the spacing
of the orbitals in the two methods.
Answer
2 2
n h
2 n x
En = =
⎛ π
, n = 1,2,3....... and ψ sin
⎞
2
n
8mL
L ⎝ L ⎠
The Orbitals of the three first levels are given by :
FMO
Huckel
E 3 =
9h 2
8mL 2 ,
α−.62β
4 h 2
E 2 =
8mL 2 ,
α+.62β
E 1 =
h 2
8mL 2 ,
α+1.62β

We note that the two set of orbitals have the same nodal structures.
However the spacing for the FMO orbital levels increases more rapidly
from n = 1 to n = 3
The minimum excitation energy is
2
2 2
h
∆E = E 3 − E = 5( )
8mL
This will correspond to -1.24 β in the Huckel theory
4.
(a) The carbene molecule CH has in its triplet state the following configuration
1
1
(1a )( 1b
) ( 1b
).
2
2 2 1 1 1
1
π
1 2
Make use of the 1a orbitals to on each carbon center in ethylene to make
a σ- bond. In the same way use the two 1b orbitals to construct th - bond
(b) Discuss in the same way the formation of two π- bonds and one σ- bond
2
in HCCH from two CH fragments each with the configuration (1 σ)
( 2σ) ( 1π)
() c Finally discuss the formation of one σ - bond in ethane from two CH fragments
4 1
with the configuration ( 1e) ( 1a)
Answer
(a)
3
1b 1
1a 1
b 2

1a 1
1a 1
1b 1 1b 1
(b)
2 2 1
The HC molecules has an electronic configuration (1 σ) ( 1π) ( 2σ)
with
three unpaired spins. The 1 πx
and 1 πy
on each HC frgment can be
used to form two π- bonds whereas the two 2σ
orbitals can be used
to form the sigma bond
1π x 1π y
2σ
1σ
H
C

1π x 1π y
2σ
2σ
1π x
1π y
() c The CH 3 fragments
4 1
have the configuration ( 1e) ( 1a)
Each can use a 1a orbital on each
fragment to form a sigma bond
1a 1
1e x
1e y

\
1a 1
1a 1